Abstract

We present sharp upper and lower generalized logarithmic mean bounds for the geometric weighted mean of the geometric and harmonic means.

1. Introduction

For 𝑝∈ℝ the generalized logarithmic mean 𝐿𝑝(𝑎,𝑏) of two positive numbers 𝑎 and 𝑏 is defined by
𝐿𝑝⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩𝑎(𝑎,𝑏)=𝑎,𝑎=𝑏,𝑝+1−𝑏𝑝+1(𝑝+1)(𝑎−𝑏)1/𝑝1,𝑝≠0,𝑝≠−1,𝑎≠𝑏,𝑒𝑏𝑏𝑎𝑎1/(𝑏−𝑎),𝑝=0,𝑎≠𝑏,𝑏−𝑎log𝑏−log𝑎,𝑝=−1,𝑎≠𝑏.(1.1)

It is well-known that 𝐿𝑝(𝑎,𝑏) is continuous and strictly increasing with respect to 𝑝∈ℝ for fixed 𝑎 and 𝑏 with 𝑎≠𝑏. In the recent past, the generalized logarithmic mean has been the subject of intensive research. In particular, many remarkable inequalities for 𝐿𝑝 can be found in the literature [1–23]. The generalized logarithmic mean has applications in convex function, economics, physics, and even in meteorology [24–27]. In [26] the authors study a variant of Jensen’s functional equation involving 𝐿𝑝, which appear in a heat conduction problem. Let 𝐴(𝑎,𝑏)=(𝑎+𝑏)/2,𝐼(𝑎,𝑏)=(1/𝑒)(𝑏𝑏/𝑎𝑎)1/(𝑏−𝑎), 𝐿(𝑎,𝑏)=(𝑏−𝑎)/(log𝑏−log𝑎), √𝐺(𝑎,𝑏)=𝑎𝑏, and 𝐻(𝑎,𝑏)=2𝑎𝑏/(𝑎+𝑏) be the arithmetic, identric, logarithmic, geometric, and harmonic means of two positive numbers 𝑎 and 𝑏 with 𝑎≠𝑏, respectively. Then it is well known thatmin{𝑎,𝑏}<𝐻(𝑎,𝑏)<𝐺(𝑎,𝑏)=𝐿−2(𝑎,𝑏)<𝐿(𝑎,𝑏)=𝐿−1(𝑎,𝑏)<𝐼(𝑎,𝑏)=𝐿0(𝑎,𝑏)<𝐴(𝑎,𝑏)=𝐿1(𝑎,𝑏)<max{𝑎,𝑏}.(1.2)

In [28–30], the authors present bounds for 𝐿 and 𝐼 in terms of 𝐺 and 𝐴.

Proposition 1.1. For all positive real numbers 𝑎 and 𝑏 with 𝑎≠𝑏, one has
𝐴1/3(𝑎,𝑏)𝐺2/31(𝑎,𝑏)<𝐿(𝑎,𝑏)<32𝐴(𝑎,𝑏)+31𝐺(𝑎,𝑏),32𝐺(𝑎,𝑏)+3𝐴(𝑎,𝑏)<𝐼(𝑎,𝑏).(1.3)

Proposition 1.2. For all positive real numbers 𝑎 and 𝑏 with 𝑎≠𝑏, we have
√√𝐺(𝑎,𝑏)𝐴(𝑎,𝑏)<1𝐿(𝑎,𝑏)𝐼(𝑎,𝑏)<21(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏))<2(𝐺(𝑎,𝑏)+𝐴(𝑎,𝑏)).(1.4)

For 𝑟∈ℝ the 𝑟th power mean 𝑀𝑟(𝑎,𝑏) of two positive numbers 𝑎 and 𝑏 is defined by𝑀𝑟⎧⎪⎨⎪⎩𝑎(𝑎,𝑏)=𝑟+𝑏𝑟21/𝑟√,𝑟≠0,𝑎𝑏,𝑟=0.(1.5)

The main properties of these means are given in [32]. Several authors discussed the relationship of certain means to 𝑀𝑟. The following sharp bounds for 𝐿, 𝐼, (𝐼𝐿)1/2, and (𝐼+𝐿)/2 in terms of power means are proved in [31, 33–37].

Proposition 1.3. For all positive real numbers 𝑎 and 𝑏 with 𝑎≠𝑏 one has
𝑀0(𝑎,𝑏)<𝐿(𝑎,𝑏)<𝑀1/3(𝑎,𝑏),𝑀2/3(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝑀log2𝑀(𝑎,𝑏),0(𝑎,𝑏)<𝐼1/2(𝑎,𝑏)𝐿1/2(𝑎,𝑏)<𝑀1/21(𝑎,𝑏),2[]𝐼(𝑎,𝑏)+𝐿(𝑎,𝑏)<𝑀1/2(𝑎,𝑏).(1.6)

The following three results were established by Alzer and Qiu in [38].

Proposition 1.4. The inequalities
𝛼𝐴(𝑎,𝑏)+(1−𝛼)𝐺(𝑎,𝑏)<𝐼(𝑎,𝑏)<𝛽𝐴(𝑎,𝑏)+(1−𝛽)𝐺(𝑎,𝑏)(1.7)
hold for all positive real numbers 𝑎 and 𝑏 with 𝑎≠𝑏 if and only if
2𝛼≤32,𝛽≥𝑒=0.73575⋅⋅⋅.(1.8)

Proposition 1.5. Let 𝑎 and 𝑏 be real numbers with 𝑎≠𝑏. If 0<𝑎,𝑏≤𝑒, then
[]𝐺(𝑎,𝑏)𝐴(𝑎,𝑏)<[]𝐿(𝑎,𝑏)𝐼(𝑎,𝑏)<[]𝐴(𝑎,𝑏)𝐺(𝑎,𝑏).(1.9)
And, if 𝑎,𝑏≥𝑒, then
[]𝐴(𝑎,𝑏)𝐺(𝑎,𝑏)<[]𝐼(𝑎,𝑏)𝐿(𝑎,𝑏)<[]𝐺(𝑎,𝑏)𝐴(𝑎,𝑏).(1.10)

Proposition 1.6. For all positive real numbers 𝑎 and 𝑏 with 𝑎≠𝑏, one has
𝑀𝑐1(𝑎,𝑏)<2(𝐿(𝑎,𝑏)+𝐼(𝑎,𝑏))(1.11)
with the best possible parameter 𝑐=log2/(1+log2)=0.40938⋯

In [39] the authors presented inequalities between the generalized logarithmic mean and the product 𝐴𝛼(𝑎,𝑏)𝐺𝛽(𝑎,b)𝐻𝛾(𝑎,𝑏) for all 𝑎,𝑏>0 with 𝑎≠𝑏 and 𝛼,𝛽>0 with 𝛼+𝛽<1.

It is the aim of this paper to give a solution to the problem: for 𝛼∈(0,1), what are the greatest value 𝑝 and the least value 𝑞, such that the inequality𝐿𝑝(𝑎,𝑏)≤𝐺𝛼(𝑎,𝑏)𝐻1−𝛼(𝑎,𝑏)≤𝐿𝑞(𝑎,𝑏)(1.12)
holds for all 𝑎,𝑏>0?

2. Main Result

Theorem 2.1. For 𝛼∈(0,1) and all 𝑎,𝑏>0, one has the following:(1)𝐿3𝛼−5(𝑎,𝑏)=𝐺𝛼(𝑎,𝑏)𝐻1−𝛼(𝑎,𝑏)=𝐿−(2/𝛼)(𝑎,𝑏) for 𝛼=2/3,(2)𝐿3𝛼−5(𝑎,𝑏)≥𝐺𝛼(𝑎,𝑏)𝐻1−𝛼(𝑎,𝑏)≥𝐿−(2/𝛼)(𝑎,𝑏) for 0<𝛼<2/3, and 𝐿3𝛼−5(𝑎,𝑏)≤𝐺𝛼(𝑎,𝑏)𝐻1−𝛼(𝑎,𝑏)≤𝐿−(2/𝛼)(𝑎,𝑏) for 2/3<𝛼<1, with equality if and only if 𝑎=𝑏, and the parameters 3𝛼−5 and −2/𝛼 in each inequality cannot be improved.

Proof. (1) If 𝛼=2/3 and 𝑎=𝑏, then (1.1) implies that 𝐿3𝛼−5(𝑎,𝑏)=𝐺𝛼(𝑎,𝑏)𝐻1−𝛼(𝑎,𝑏)=𝐿−(2/𝛼)(𝑎,𝑏)=𝑎.If 𝛼=2/3 and 𝑎≠𝑏, then (1.1) leads to
𝐿3𝛼−5(𝑎,𝑏)=𝐿−2/𝛼(𝑎,𝑏)=𝐿−3𝑎(𝑎,𝑏)=−2−𝑏−22(𝑏−𝑎)−1/3=(𝑎𝑏)1/32𝑎𝑏𝑎+𝑏1/3=𝐺2/3(𝑎,𝑏)𝐻1/3(𝑎,𝑏)=𝐺𝛼(𝑎,𝑏)𝐻1−𝛼(𝑎,𝑏).(2.1) (2) If 𝑎=𝑏, then from (1.1) we clearly see that 𝐿3𝛼−5(𝑎,𝑏)=𝐺𝛼(𝑎,𝑏)𝐻1−𝛼(𝑎,𝑏)=𝐿−(2/𝛼)(𝑎,𝑏)=𝑎 for any 𝛼∈(0,1). If 𝑎≠𝑏, without loss of generality, we assume 𝑎>𝑏. Let 𝑎/𝑏=𝑡>1 and
𝑓(𝑡)=log𝐿3𝛼−5𝐺(𝑎,𝑏)−log𝛼(𝑎,𝑏)𝐻1−𝛼.(𝑎,𝑏)(2.2)
Then (1.1) and simple computations yield
1𝑓(𝑡)=𝑡3𝛼−5log3𝛼−4−1−𝛼(3𝛼−4)(𝑡−1)2log𝑡−(1−𝛼)log2𝑡,1+𝑡lim𝑡→1+𝑓𝑓(𝑡)=0,(2.3)𝑡(𝑡)=−4−3𝛼𝑡𝑡2𝑡−14−3𝛼−1𝑔(𝑡),(2.4)
where 𝑔(𝑡)=(2−𝛼/2)𝑡3𝛼−2−((2−𝛼)(2−3𝛼)/5−3𝛼)𝑡3𝛼−3+((1−𝛼)(2−3𝛼)/2(5−3𝛼))𝑡3𝛼−4−((1−𝛼)(2−3𝛼)/2(5−3𝛼))𝑡2+((2−𝛼)(2−3𝛼)/(5−3𝛼))𝑡−(2−𝛼)/2,
𝑔𝑔(1)=0,(𝑡)=(2−𝛼)(3𝛼−2)2𝑡3𝛼−3−3(2−𝛼)(2−3𝛼)(𝛼−1)𝑡5−3𝛼3𝛼−4+(1−𝛼)(2−3𝛼)(3𝛼−4)𝑡2(5−3𝛼)3𝛼−5−(1−𝛼)(2−3𝛼)𝑡+(5−3𝛼)(2−𝛼)(2−3𝛼),𝑔(5−3𝛼)𝑔(1)=0,(𝑡)=3(2−𝛼)(3𝛼−2)(𝛼−1)2𝑡3𝛼−4−3(2−𝛼)(2−3𝛼)(𝛼−1)(3𝛼−4)𝑡5−3𝛼3𝛼−5−(1−𝛼)(2−3𝛼)(3𝛼−4)2𝑡3𝛼−6−(1−𝛼)(2−3𝛼),𝑔(5−3𝛼)(2.5)𝑔(1)=0,(2.6)3(𝑡)=2(1−𝛼)(2−𝛼)(4−3𝛼)(3𝛼−2)𝑡3𝛼−7(𝑡−1)2.(2.7)If 0<𝛼<2/3, then (2.7) implies
𝑔(𝑡)<0(2.8)
for 𝑡>1.From (2.3)–(2.6) and (2.8) we know that 𝑓(𝑡)>0 for 𝑡>1.If 2/3<𝛼<1, then (2.7) leads to
𝑔(𝑡)>0(2.9)
for 𝑡>1. Therefore 𝑓(𝑡)<0 for 𝑡>1 follows from (2.3)–(2.6) and (2.9).Let
ℎ(𝑡)=log𝐿−(2/𝛼)𝐺(𝑎,𝑏)−log𝛼(𝑎,𝑏)𝐻1−𝛼(𝑎,𝑏)(2.10)
for 𝑡=𝑎/𝑏>1; then (1.1) and elementary calculations lead to
𝛼ℎ(𝑡)=−2𝑡log(𝛼−2)/𝛼−1−𝛼((𝛼−2)/𝛼)(𝑡−1)2log𝑡−(1−𝛼)log2𝑡,1+𝑡lim𝑡→1+ℎℎ(𝑡)=0,(2.11)𝑡(𝑡)=−(2−𝛼)/𝛼𝑡𝑡2𝑡−1(2−𝛼)/𝛼−1𝑣(𝑡),(2.12)
where 𝑣(𝑡)=((2−𝛼)/2)𝑡(3𝛼−2)/𝛼+((3𝛼−2)/2)𝑡(2𝛼−2)/𝛼−((3𝛼−2)/2)𝑡−(2−𝛼)/2,
𝑣𝑣(1)=0,(𝑡)=(2−𝛼)(3𝛼−2)𝑡2𝛼(2𝛼−2)/𝛼+(3𝛼−2)(𝛼−1)𝛼𝑡(𝛼−2)/𝛼−3𝛼−22,𝑣(2.13)𝑣(1)=0,(2.14)(𝑡)=(2−𝛼)(1−𝛼)(2−3𝛼)𝛼2𝑡−2/𝛼(𝑡−1).(2.15)If 𝛼∈(0,2/3), then (2.15) implies
𝑣(𝑡)>0(2.16)
for 𝑡>1.From (2.11)–(2.14) and (2.16) we know that ℎ(𝑡)<0 for 𝑡>1.If 𝛼∈(2/3,1), then (2.15) leads to
𝑣(𝑡)<0(2.17)
for 𝑡>1. Therefore, ℎ(𝑡)>0 for 𝑡>1 follows from (2.11)–(2.14) and (2.17).Next, we prove that the parameters −(2/𝛼) and 3𝛼−5 in either case cannot be improved. The proof is divided into two cases.Case 1 (𝛼∈(0,2/3)). For any 𝜖>0 and 𝑥∈(0,1), from (1.1) one has
𝐺𝛼(1,1+𝑥)𝐻1−𝛼(1,1+𝑥)5−3𝛼+𝜖−𝐿3𝛼−5−𝜖(1,1+𝑥)5−3𝛼+𝜖=𝑓1(𝑥)(1+𝑥/2)(1−𝛼)(5−3𝛼+𝜖)(1+𝑥)4−3𝛼+𝜖,−1(2.18)
where 𝑓1(𝑥)=(1+𝑥)(1−𝛼/2)(5−3𝛼+𝜖)[(1+𝑥)4−3𝛼+𝜖−1]−(4−3𝛼+𝜖)𝑥(1+𝑥)4−3𝛼+𝜖(1+𝑥/2)(1−𝛼)(5−3𝛼+𝜖).Let 𝑥→0; making use of the Taylor expansion, we get
𝑓1(𝑥)=𝜖(4−3𝛼+𝜖)(5−3𝛼+𝜖)𝑥243𝑥+𝑜3.(2.19)Equations (2.18) and (2.19) imply that for any 𝛼∈(0,2/3) and 𝜖>0 there exists 𝛿=𝛿(𝜖,𝛼)∈(0,1), such that 𝐿3𝛼−5−𝜖(1,1+𝑥)<𝐺𝛼(1,1+𝑥)𝐻1−𝛼(1,1+𝑥) for 𝑥∈(0,𝛿).On the other hand, for any 𝜖∈(0,(2/𝛼)−1) we have
𝐿−(2/𝛼)+𝜖(1,𝑡)−𝐺𝛼(1,𝑡)𝐻1−𝛼(1,𝑡)=𝑡𝛼/(2−𝜖𝛼)1−𝑡−2/𝛼+𝜖+1(2/𝛼−𝜖−1)(1−1/𝑡)−𝛼/(2−𝜖𝛼)−𝑡−𝜖𝛼2/2(2−𝜖𝛼)2𝑡1+𝑡1−𝛼,lim𝑡→+∞1−𝑡−2/𝛼+𝜖+1(2/𝛼−𝜖−1)(1−1/𝑡)−𝛼/(2−𝜖𝛼)−𝑡−𝜖𝛼2/2(2−𝜖𝛼)2𝑡1+𝑡1−𝛼=2𝛼−𝜖−1𝛼/(2−𝜖𝛼)>0.(2.20)From (2.20) we know that for any 𝛼∈(0,2/3) and 𝜖∈(0,2/𝛼−1) there exists 𝑇=𝑇(𝜖,𝛼)>1, such that 𝐿−2/𝛼+𝜖(1,𝑡)>𝐺𝛼(1,𝑡)𝐻1−𝛼(1,𝑡) for 𝑡∈(𝑇,∞).Case 2 (𝛼∈(2/3,1)). For any 𝜖∈(0,4−3𝛼) and 𝑥∈(0,1), from (1.1) one has
𝐿3𝛼−5+𝜖(1,1+𝑥)5−3𝛼−𝜖−𝐺𝛼(1,1+𝑥)𝐻1−𝛼(1,1+𝑥)5−3𝛼−𝜖=𝑓2(𝑥)(1+𝑥/2)(1−𝛼)(5−3𝛼−𝜖)(1+𝑥)4−3𝛼−𝜖,−1(2.21)
where 𝑓2(𝑥)=(4−3𝛼−𝜖)𝑥(1+𝑥)4−3𝛼−𝜖(1+𝑥/2)(1−𝛼)(5−3𝛼−𝜖)−(1+𝑥)(1−𝛼/2)(5−3𝛼−𝜖)[(1+𝑥)4−3𝛼−𝜖−1].Let 𝑥→0; making use of the Taylor expansion, we have
𝑓2𝜖(𝑥)=24(4−3𝛼−𝜖)(5−3𝛼−𝜖)𝑥3𝑥+𝑜3.(2.22)Equations (2.21) and (2.22) imply that for any 𝛼∈(2/3,1) and 𝜖∈(0,4−3𝛼) there exists 𝛿=𝛿(𝜖,𝛼)∈(0,1), such that 𝐿3𝛼−5+𝜖(1,1+𝑥)>𝐺𝛼(1,1+𝑥)𝐻1−𝛼(1,1+𝑥) for 𝑥∈(0,𝛿).On the other hand, for any 𝜖>0, we have
𝐺𝛼(1,𝑡)𝐻1−𝛼(1,𝑡)−𝐿−(2/𝛼)−𝜖(1,𝑡)=𝑡𝛼/22𝑡1+𝑡1−𝛼−𝑡−𝜖𝛼2/2(2+𝜖𝛼)1−𝑡−(2/𝛼+𝜖−1)(2/𝛼+𝜖−1)(1−1/𝑡)−𝛼/(2+𝜖𝛼),lim𝑡→+∞2𝑡1+𝑡1−𝛼−𝑡−𝜖𝛼2/2(2+𝜖𝛼)1−𝑡−(2/𝛼+𝜖−1)(2/𝛼+𝜖−1)(1−1/𝑡)−𝛼/(2+𝜖𝛼)=21−𝛼>0.(2.23)From (2.23) we know that for any 𝛼∈(2/3,1) and 𝜖>0 there exists 𝑇=𝑇(𝜖,𝛼)>1, such that 𝐿−(2/𝛼)−𝜖(1,𝑡)<𝐺𝛼(1,𝑡)𝐻1−𝛼(1,𝑡) for 𝑡∈(𝑇,∞).

Acknowledgment

This work was supported by the Natural Science Foundation of Zhejiang Broad-cast and TV University under Grant XKT-09G21.